Answer:
Butanoic Acid:
I have attached the structure of Butanoic Acid. Please refer to the attachment.
Explanation:
Solution:
First of all, we need to calculate the double bond equivalent (DBE) for the compound given:
Compound = C4H8O2
DBE = (C +1) - (H + X - N)/2
C = no. of carbons
H = no. of Hydrogen
X = no. of Halogens
N = no. of Nitrogen
DBE = (4+1) - (8/2)
DBE = 5-4
DBE = 1
It indicates the presence of 1 double bond in the possible structure.
Further structure can be identified by looking the peak values of the spectrum. So, we have,
2500 - 3300 = it indicates the presence of -COOH group
1720 = it indicates the (C=0) stretching
1200 = It indicates the (C-O) stretching
SO, the possible structure is
Butanoic Acid:
I have attached the structure of Butanoic Acid. Please refer to the attachment.
Ice, freezing, or melting then liquid evaporation or condensation, the vapor, deposition the ice again
Hydrogen-3 with scientific name Tritium often undergoes beta decay.
I believe such reactions are called decomposition reactions. Decomposition reactions are reactions that involve breaking down a compound into smaller compounds or individual elements. They involve input of energy in the form of heat, electricity or light to break down the compounds. When light is used the reaction is referred to as photo decomposition while when heat is used the reaction is thermal decomposition reaction e.t.c. The general form of the reaction is AB = A + B.
Answer:
C. 30.0 g NaOH and add water until the final solution has a volume of 3.00 L.
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
from the question ,
M = 0.250M
V = 3.00 L
M = n / V
n = M * v
n = 0.250M * 3.00 L = 0.75 mol
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
n = 0.75 mol NaOH
m = molecular mass of NaOH = 40 g/mol
n = w / m
w = n * m
w = 0.75 mol * 40 g/mol = 30.0 g
Hence , by using 30.0 g of NaOH and dissolving it to make up the volume to 3 L , a solution of 0.250 M can be prepared .
